New Notes Blog: `int sqrt(25x^2+4)/x^4 dx` Find the indefinite integral

Saturday, 22 February 2014

$\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

To solve the indefinite integral, we follow $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as the constant of integration.

For the given integral problem: $\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ , we may apply integration by parts: $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}\cdot{d}{u}$ .

Let:

$\displaystyle{u}=\sqrt{{{25}{{x}}^{{2}}+{4}}}$

Apply Law of Exponent: $\displaystyle\sqrt{{{x}}}={{x}}^{{\frac{{1}}{{2}}}}$ , we get: $\displaystyle{u}={{\left({25}{{x}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}$

To find the derivative of $\displaystyle{u}$ , we may apply Power rule for derivative: $\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{{u}}^{{n}}={n}\cdot{{u}}^{{{n}-{1}}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({u}\right)}$

$\displaystyle{u}'=\frac{{1}}{{2}}\cdot{{\left({25}{{x}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}-{1}}}\cdot\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({25}{{x}}^{{2}}+{4}\right)}$

$\displaystyle{u}'=\frac{{1}}{{2}}\cdot{{\left({25}{{x}}^{{2}}+{4}\right)}}^{{-\frac{{1}}{{2}}}}\cdot{\left({50}{x}{\left.{d}{x}\right.}\right)}$

$\displaystyle{u}'={25}{x}{{\left({25}{{x}}^{{2}}+{4}\right)}}^{{-\frac{{1}}{{2}}}}{\left.{d}{x}\right.}$

Apply Law of exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ .

$\displaystyle{u}'=\frac{{{25}{x}}}{{{\left({25}{{x}}^{{2}}+{4}\right)}}^{{\frac{{1}}{{2}}}}}{\left.{d}{x}\right.}{\quad\text{or}\quad}\frac{{{25}{x}}}{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{\left.{d}{x}\right.}$

Let: $\displaystyle{v}'=\frac{{1}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$

To find the integral of $\displaystyle{v}'$ , we apply Law of exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ and Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle{v}=\int{v}'$

$\displaystyle=\int\frac{{1}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$

$\displaystyle=\int{{x}}^{{-{4}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{-{4}+{1}}}}{{-{4}+{1}}}$

$\displaystyle=\frac{{{x}}^{{-{3}}}}{{-{3}}}$

$\displaystyle=-\frac{{1}}{{{3}{{x}}^{{3}}}}$

Apply the formula for integration by parts using the following values: $\displaystyle{u}=\sqrt{{{25}{{x}}^{{2}}+{4}}}$ , $\displaystyle{u}'=\frac{{{25}{x}}}{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{\left.{d}{x}\right.}$ , $\displaystyle{v}'=\frac{{1}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ and $\displaystyle{v}=-\frac{{1}}{{{3}{{x}}^{{3}}}}$ .

$\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}=\sqrt{{{25}{{x}}^{{2}}+{4}}}\cdot-{\left(\frac{{1}}{{{3}{{x}}^{{3}}}}\right)}-\int\frac{{{25}{x}}}{\sqrt{{{25}{{x}}^{{2}}+{4}}}}\cdot{\left(-\frac{{1}}{{{3}{{x}}^{{3}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{3}{{x}}^{{3}}}}-\int-\frac{{25}}{{{3}{{x}}^{{2}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}$

To evaluate the integral part, we may apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int-\frac{{{25}{x}}}{{{3}{{x}}^{{3}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}=-\frac{{25}}{{3}}\int\frac{{1}}{{{{x}}^{{2}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}$

The integral resembles one of the formulas from the integration table for rational function with roots. We follow:

$\displaystyle\int\frac{{{d}{u}}}{{{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{{a}}^{{2}}}}}}=-\frac{\sqrt{{{{u}}^{{2}}+{{a}}^{{2}}}}}{{{{a}}^{{2}}{x}}}+{C}$

For easier comparison, we may apply u-substitution by letting: $\displaystyle{{u}}^{{2}}={25}{{x}}^{{2}}$ or $\displaystyle{{\left({5}{x}\right)}}^{{2}}$ then $\displaystyle{u}={5}{x}$ and $\displaystyle{d}{u}={5}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{5}}={\left.{d}{x}\right.}$ . When we let $\displaystyle{{u}}^{{2}}={25}{{x}}^{{2}}$ , it can be rearrange as $\displaystyle{{x}}^{{2}}=\frac{{{u}}^{{2}}}{{25}}$ . Applying the values, the integral becomes:

$\displaystyle-\frac{{25}}{{3}}\int\frac{{1}}{{{{x}}^{{2}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}=-\frac{{25}}{{3}}\int\frac{{1}}{{{\left(\frac{{{u}}^{{2}}}{{25}}\right)}\sqrt{{{{u}}^{{2}}+{4}}}}}\cdot\frac{{{d}{u}}}{{5}}$

$\displaystyle=-\frac{{25}}{{3}}\int\frac{{25}}{{{5}{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{4}}}}}{d}{u}$

$\displaystyle=-\frac{{125}}{{3}}\int\frac{{1}}{{{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}$

By comparing " $\displaystyle{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{{a}}^{{2}}}}$ " with " $\displaystyle{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{4}}}$ ", we determine the corresponding value: $\displaystyle{{a}}^{{2}}={4}$ . Applying the aforementioned integration formula for rational function with roots, we get:

$\displaystyle-\frac{{125}}{{3}}\int\frac{{1}}{{{{u}}^{{2}}\sqrt{{{{u}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}=-\frac{{125}}{{3}}\cdot{\left[-\frac{\sqrt{{{{u}}^{{2}}+{4}}}}{{{4}{u}}}\right]}+{C}$

$\displaystyle=\frac{{{125}\sqrt{{{{u}}^{{2}}+{4}}}}}{{{12}{u}}}+{C}$

Plug-in $\displaystyle{{u}}^{{2}}={25}{{x}}^{{2}}$ and $\displaystyle{u}={5}{x}$ on $\displaystyle\frac{{{125}\sqrt{{{{u}}^{{2}}+{4}}}}}{{{12}{u}}}+{C}$ , we get the indefinite integral:

$\displaystyle\int-\frac{{25}}{{{3}{{x}}^{{2}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}=\frac{{{125}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{{{12}\cdot{5}{x}}}+{C}$

$\displaystyle=\frac{{{25}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{{{12}{x}}}+{C}$ .

For the complete indefinite integral, we get:

$\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}=-\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{3}{{x}}^{{3}}}}-\int-\frac{{25}}{{{3}{{x}}^{{2}}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{\left.{d}{x}\right.}$

$\displaystyle=-\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{3}{{x}}^{{3}}}}-\frac{{{25}\sqrt{{{25}{{x}}^{{2}}+{4}}}}}{{{12}{x}}}+{C}$

New Notes Blog

Saturday, 22 February 2014

$\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 22 February 2014

Find the indefinite integral

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In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{\sqrt{{{25}{{x}}^{{2}}+{4}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?